Question

# The standard heat of formation, ΔH∘f, is defined as the enthalpy change for the formation of...

The standard heat of formation, ΔH∘f, is defined as the enthalpy change for the formation of one mole of substance from its constituent elements in their standard states. Thus, elements in their standard states have ΔH∘f=0. Heat of formation values can be used to calculate the enthalpy change of any reaction.

Consider, for example, the reaction

2NO(g)+O2(g)⇌2NO2(g)

with heat of formation values given by the following table:

 Substance ΔH∘f (kJ/mol) NO(g) 90.2 O2(g) 0 NO2(g) 33.2

Then the standard heat of reaction for the overall reaction is

ΔH∘rxn===ΔH∘f(products)2(33.2)−114 kJ−−ΔH∘f(reactants)[2(90.2)+0]

For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)?

You do not need to look up any values to answer this question.

Check all that apply.

Hints

Check all that apply.

S(s)+O2(g)→SO2(g)
Li(s)+12Cl2(g)→LiCl(s)
SO3(g)→12O2(g)+SO2(g)
Li(s)+12Cl2(l)→LiCl(s)
2Li(s)+Cl2(g)→2LiCl(s)

SO(g)+12O2(g)→SO2(g)

The combustion of ethene, C2H4, occurs via the reaction

C2H4(g)+3O2(g)→2CO2(g)+2H2O(g)

with heat of formation values given by the following table:

 Substance ΔH∘f (kJ/mol) C2H4 (g) 52.47 CO2(g) −393.5 H2O(g) −241.8

Calculate the enthalpy for the combustion of 1 mole of ethene.

HRxn = Hproducts - Hreactants

HRxn = (2NO2) - (2NO + O2)

HRxn = (2*33.2) - (2*90.2 + 0)

HRxn = -114 kJ/mol

Q2.

For

heat of formation of a species:

- Must be elemental state of al of its elements

- Miust be the most stable form at STP

- Must be 1 mol of product produced

therefore:

S(s)+O2(g)→SO2(g) This is Hformation,

Li(s)+1/2Cl2(g)→LiCl(s) --> Correct, it form sa single mol of LiCl

SO3(g)→12O2(g)+SO2(g) --> false, it is forming more than 1 mol

Li(s)+1/2Cl2(l)→LiCl(s) --> false, Chlorine is not liquid at STP

2Li(s)+Cl2(g)→2LiCl(s) --> false, it must produce only 1 mol

SO(g)+12O2(g)→SO2(g) --> False, SO is not elemental

Q#.

HComb = Hproducts - Hreactants

Hcomb = (2*CO2 + 2H2O) - (C2H4 + 3O2)

HRcomb = (2*-393.5 + 2*-241.8) - (52.47 + 3*0)

Hcomb = -1323.07 kJ/mol